Divergence of a series (Raabe fails)

Question

Consider $\displaystyle a_n=\left(\frac{(2n)!}{2^{2n}(n!)^2}\right)^2$

Prove that $\sum a_n$ diverges

Lots of factorials, so first thing is to check for ratio test (fails), Raabe test (also fails).

I can't find any lower bound that goes to infinity...

By the Stirling approximation, $a_n$ behaves like $1/n$. – André Nicolas Jun 15 '14 at 15:17 — André Nicolas, Jun 15 '14 at 15:17
@AndréNicolas do you wish to make it an answer ? – Gabriel Romon Jun 15 '14 at 15:29 — Gabriel Romon, Jun 15 '14 at 15:29

score 4 · Accepted Answer · answered Jun 15 '14 at 15:38

We have the Stirling approximation $$n!\sim_\infty \left(\frac ne\right)^n\sqrt{2\pi n}$$ hence

$$\frac{(2n)!}{2^{2n}(n!)^2}\sim_\infty\frac{(2n)^{2n}\sqrt{4\pi n}}{2^{2n} n^{2n}\times2\pi n}=\frac{1}{\sqrt{\pi n}}$$ hence we have $$a_n\sim_\infty \frac1{\pi n}$$ and then the series is divergent by comparison with the harmonic series.

Divergence of a series (Raabe fails)

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